How does intensity affect the photoelectric effect?

It increases or decreases the number of electrons ejected from the metal surface.

So, the intensity of light (number of photons) incident on the metal surface has no effect on the energy of the ejected electrons. This can be understood by the photoelectric equation :

##color(white)(11sds)hnu=phi+KE_max##

##or, KE_max=hnu-phi##

where, ##KE_max->## maximum Kinetic Energy of ejected electron ##hcolor(white)11111->## ##nucolor(white)11111->##frequency of radiation and ##phicolor(white)11111->##work function of the metal

From this equation, it is clear that the number of photons incident has no effect on the energy of the photoelectrons

However, the the intensity of light decides the number of photoelectrons that are released. As each photon is capable of releasing only one electron, then if there are more photons, then there is release of more photons.

Lets suppose that if ##100## photons are incident, then about ##X## electrons are emitted. If we increase the number of Photons (i.e. increase the intensity of light) to lets say ##1000##, then the number of photo electrons will be significantly higher than ##X##